In a previous post, we were introduced to

Pascal's Hexagrammum Mysticum Theorem - a magical theorem - which states that if we draw a

*hexagon* inscribed in a

*conic section* then the three pairs of

*opposite sides* of the hexagon intersect at three points which lie on a

*straight line*.

For example, as in the following figure we have a hexagon inscribed in a

*circle* and the intersection points of the three pairs of the opposite sides of the hexagon $\{12, 45\}$, $\{23, 56\}$, $\{34, 61\}$ are collinear.

There is a useful tool to prove the

*collinearity of points* - the

Menelaus' Theorem - which states as follows:

**Menelaus' Theorem:** Given a triangle $ABC$ and three points $A'$, $B'$, $C'$ lying on the three *lines* $BC$, $CA$, $AB$, respectively. Then the three points $A'$, $B'$, $C'$ are **collinear** if and only if $$\frac{\vec{A'B}}{\vec{A'C}} \times \frac{\vec{B'C}}{\vec{B'A}} \times \frac{\vec{C'A}}{\vec{C'B}} = 1.$$

Today, we will use

*Menelaus' theorem* to prove

* Pascal's theorem* for the circle case.